# Sunday Afternoon Maths XXXI

## Integrals

$$\int_0^1 1 dx = 1$$

Find \(a_1\) such that:

$$\int_0^{a_1} x dx = 1$$

Find \(a_2\) such that:

$$\int_0^{a_2} x^2 dx = 1$$

Find \(a_n\) such that (for \(n>0\)):

$$\int_0^{a_n} x^n dx = 1$$

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$$1=\int_0^{a_1} x dx=\frac{a_1^2}{2}$$

So, \(a_1=\sqrt{2}\).

$$1=\int_0^{a_2} x^2 dx=\frac{a_2^3}{3}$$

So, \(a_2=3^{\frac{1}{3}}\).

$$1=\int_0^{a_n} x^n dx=\frac{a_n^{n+1}}{n+1}$$

So, \(a_n={(n+1)}^{\frac{1}{n+1}}\).

#### Extension

Find \(b_n\) such that (for \(n>1\)):

$$\int_{b_n}^{\infty} x^{-n} dx = 1$$

## Tetrahedral die

When a tetrahedral die is rolled, it will land with a point at the top: there is no upwards face on which the value of the roll can be printed. This is usually solved by printing three numbers on each face and the number which is at the bottom of the face is the value of the roll.

Is it possible to make a tetrahedral die with one number on each face such that the value of the roll can be calculated by adding up the three visible numbers? (the values of the four rolls must be 1, 2, 3 and 4)

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Let \(a\), \(b\), \(c\) and \(d\) be the numbers on the four faces. The following simultaneous equations must hold:

$$a+b+c=1$$
$$a+b+d=2$$
$$a+c+d=3$$
$$b+c+d=4$$

These can be solved to find that the numbers on the faces must be \(-\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{4}{3}\) and \(\frac{7}{3}\).

#### Extension

Is it possible to make a six-sided die with one number on each face such that the value of the roll can be calculated by adding up the five visible numbers?

Is it possible to make an \(n\)-sided die with one number on each face such that the value of the roll can be calculated by adding up the \((n-1)\) visible numbers?

Is it possible to make a die with one **integer** on each face such that the value of the roll can be calculated by adding up the visible numbers?